MATH 110: LINEAR ALGEBRA
SPRING 2007/08
PROBLEM SET 8 SOLUTIONS
1.
Let
V
be a vector space over
F
and
S
,
T ∈
End(
V
).
(a) Show that
I  S ◦ T
is injective iff
I  T ◦ S
is injective.
Solution.
The trick is to observe that
S ◦
(
I  T ◦ S
) = (
I  S ◦ T
)
◦ S
.
Suppose
I  S ◦ T
is injective. Let
v
∈
ker(
I  T ◦ S
). Then (
I  T ◦ S
)(
v
) =
0
V
. Hence
(
I  S ◦ T
)(
S
(
v
)) =
S
((
I  T ◦ S
)(
v
)) =
S
(
0
V
) =
0
V
and since
I  S ◦ T
is injective, we deduce that
S
(
v
) =
0
V
.
So
T ◦ S
(
v
) =
T
(
0
V
) =
0
V
;
and so
v
=
v

0
V
=
v
 T ◦ S
(
v
) = (
I  T ◦ S
)(
v
) =
0
V
where the last equality follows from our choice of
v
∈
ker(
I T ◦S
). Hence ker(
I T ◦S
) =
{
0
V
}
and we deduce that
I  T ◦ S
is injective by Theorem
4.12
.
(b)
T
is called
nilpotent
if
T
n
=
O
for some
n
∈
N
. Show that if
T
is nilpotent, then
I  T
is bijective. What is (
I  T
)

1
?
Solution.
Let
S
=
I
+
T
+
T
2
+
· · ·
+
T
n

1
(by the geometric series heuristic in lecture).
Then
(
I  T
)
◦ S
= (
I
+
T
+
· · ·
+
T
n

1
)

(
T
+
T
2
+
· · ·
+
T
n
) =
I  T
n
=
I  O
=
I
and likewise for
S ◦
(
I  T
). Hence
(
I  T
)

1
=
S
and
I  T
is bijective.
2.
Let
V
be a vector space over
R
and
T ∈
End(
V
) be an involution, ie.
T
2
=
I
. Define
V
+
:=
{
v
∈
V
 T
(
v
) =
v
}
and
V

:=
{
v
∈
V
 T
(
v
) =

v
}
.
(a) Show that
V
+
and
V

are subspaces of
V
.
Solution.
Let
α
1
, α
2
∈
R
and
v
1
,
v
2
∈
V
+
. Then
T
(
v
1
) =
v
1
,
T
(
v
2
) =
v
2
, and
T
(
α
1
v
1
+
α
2
v
2
) =
α
1
T
(
v
1
) +
α
2
T
(
v
2
)
=
α
1
v
1
+
α
2
v
2
.
Hence
α
1
v
1
+
α
2
v
2
∈
V
+
. Likewise for
V

.
Date
: May 7, 2008 (Version 1.0).
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(b) Show that
V
+
⊕
V

=
V.
Solution.
Note that
v
=
1
2
(
I
+
T
)(
v
) +
1
2
(
I  T
)(
v
)
.
(2.1)
Since
T
2
=
I
,
T
(
1
2
(
I
+
T
)(
v
)
)
=
T
(
1
2
v
+
1
2
T
(
v
)
)
=
1
2
T
(
v
) +
1
2
T
2
(
v
) =
1
2
T
(
v
) +
1
2
I
(
v
) =
1
2
(
I
+
T
)(
v
)
and likewise,
T
(
1
2
(
I  T
)(
v
)
)
=
T
(
1
2
v

1
2
T
(
v
)
)
=
1
2
T
(
v
)

1
2
T
2
(
v
) =
1
2
T
(
v
)

1
2
I
(
v
) =

1
2
(
I  T
)(
v
)
.
Hence
1
2
(
I
+
T
)(
v
)
∈
V
+
,
1
2
(
I  T
)(
v
)
∈
V

,
and (2.1) implies that
V
=
V
+
+
V

. If
v
∈
V
+
∩
V

, then
v
=
T
(
v
) =

v
implies that
v
=
0
V
. Hence
V
+
∩
V

=
{
0
V
}
and so
V
=
V
+
⊕
V

.
3.
Let
A
∈
R
m
×
n
and
b
∈
R
m
.
(a) Show that
nullsp(
A
>
A
) = nullsp(
A
)
.
Solution.
Let
x
∈
nullsp(
A
). Then
A
x
=
0
. Therefore,
A
>
A
x
=
A
>
0
=
0
. Therefore
x
∈
nullsp(
A
>
A
). So nullsp(
A
)
⊆
nullsp(
A
>
A
). Conversely, let
x
∈
nullsp(
A
>
A
). Then
A
>
A
x
=
0
. Multiplying on the left by
x
>
, we get
x
>
A
>
A
x
=
x
>
0
= 0
.
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 Spring '08
 GUREVITCH
 Math, Linear Algebra, Algebra, Vector Space, wi wi

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